Chapter 5.2 — Exercise 5.1 — Formation

Eight Equations, Simplified First

Every equation here needs expanding and simplifying before its true degree becomes visible — none of the eight can be judged correctly from its starting, unsimplified appearance alone, which is exactly the point of including them in this exercise.

The last two rows make an instructive pair: both start out looking cubic on either side, but only one of them actually stays cubic after expanding. In the seventh equation, (x+2)³ = 2x(x²−1), expanding the left side gives x³+6x²+12x+8 and the right side gives 2x³−2x, and subtracting one full expansion from the other leaves −x³+6x²+14x+8=0 — the x³ terms don't cancel, so degree 3 survives. In the eighth equation, x³−4x²−x+1=(x−2)³, expanding the cube on the right-hand side (x−2)³ gives x³−6x²+12x−8, and subtracting that from the left side's x³−4x²−x+1 cancels the x³ terms completely, leaving 2x²−13x+9=0 — genuinely quadratic. Full expansion, not a quick glance at the highest power written down before simplifying, is the only reliable test for either case.

A Rectangular Plot, Described Algebraically

A rectangular plot has area 528 m², and its length is one metre more than twice its breadth. Naming the unknown breadth x m makes the length automatically 2x + 1 m, expressed in terms of the same single variable, so:

This is exactly what "formation" means throughout this exercise: translate the relationship between two unknowns into one equation in a single variable, using the given relationship to eliminate the second unknown entirely before it's ever separately introduced. Notice that forming the equation never actually required solving it — the whole point of this exercise is the translation step alone, leaving 2x² + x − 528 = 0 sitting there unsolved, ready for whichever solving method the next exercise introduces.

Three More Situations, Formed the Same Way

• Two consecutive positive integers whose product is 306 — naming the smaller one x makes the larger one automatically x+1, with no second unknown to introduce separately. Their product gives x(x+1) = 306, which expands to x² + x − 306 = 0.
• Rohan's age problem: his mother is 26 years older, and in 3 years the product of their ages will be 360. Naming Rohan's present age x, his mother's present age is automatically x+26 — the "26 years older" clue does the same job the "+1" did for consecutive integers above. Three years from now, Rohan will be x+3 and his mother will be x+26+3 = x+29, so (x+3)(x+29) = 360 expands to x² + 32x − 273 = 0.
• A train travelling 480 km: at 8 km/h slower, the journey takes 3 hours longer. Naming the actual speed x kmph, the time taken is 480/x hours; at the reduced speed x−8, the time taken is 480/(x−8) hours, and the difference between these two times is exactly 3 hours: 480/(x−8) − 480/x = 3, which clears to x² − 8x − 1280 = 0.

The train problem is worth slowing down on, since it's the only one of the four that starts as a fraction rather than a product. Multiplying both sides by x(x−8) — the product of both denominators — clears the fractions in one step and leaves a standard quadratic behind; skipping straight to that multiplication, rather than trying to combine the fractions piece by piece, is the fastest route through this kind of setup. It's also genuinely worth pausing to notice what the resulting equation does not yet tell you: x²−8x−1280=0 is a fully valid quadratic equation, but nothing about its formation has determined which of its (as yet unknown) roots is the actual train speed — that question belongs entirely to the solving methods covered in the exercises still ahead in this same chapter.

A Pattern Worth Naming

Every situation in this exercise reduces to the same shape: two unknowns linked by one stated relationship, which lets a single chosen variable stand in for both quantities at once. Once that link is written down algebraically — "one more than twice," "26 years older," "8 km/h slower" — the rest is just multiplying out and collecting terms into ax² + bx + c = 0 form. Recognising this shared pattern is what turns a wordy paragraph into a routine translation exercise rather than a fresh puzzle to be solved from first principles each time.

Building a mental library of phrases like these — and the single variable expression each one becomes — is what makes formation faster over time. The vocabulary changes from problem to problem, but the underlying translation move is identical every time: find the one unknown that everything else in the problem can be expressed in terms of, then write down the relationship the problem actually states, in exactly the words it was given, without adding or losing any information along the way.

From Forming to Solving

Every equation formed in this exercise is left unsolved — writing it down correctly was the whole point, and the payoff for getting the setup right shows up in the next exercise, not this one. Exercise 5.2 picks up exactly where this leaves off, introducing factorisation as the first method for actually finding the roots of equations shaped just like the plot, integer, age, and train problems formed here — geometry, integer, age, and rate word problems of the same general kind, now solved all the way through to real numeric answers. For the standard-form definitions this exercise assumes throughout, revisit the chapter introduction.